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Answer 1 · 2017-05-20T16:56:50

$2.6 "g Fe"_2"O"_3$

The reaction for this problem is:

$4"Fe"_3"O"_4 + "O"_2 -> 6"Fe"_2"O"_3$

The molar mass of $"Fe"_3"O"_4$ is 231.533 g/mol

So, you have $2.5 "g" times (1" mol")/(231.533"g") = 0.0108" mol " "Fe"_3"O"_4$

Now multiply this by the $6:4$ ratio of $"Fe"_2"O"_3$ to $"Fe"_3"O"_4$ to get $0.0162 " mol ""Fe"_2"O"_3$ .

Finally multiply this by the molar mass of $"Fe"_2"O"_3 - 159.69 " g/mol"$ .

$0.0162 " mol" times 159.69 " g/mol" = 2.6 "g Fe"_2"O"_3$

Final Answer